Tuesday, August 24, 2021

H C Verma solutions, MAGNETIC FIELD, Chapter-34, Exercises, Q31 -Q40, Concepts of Physics, Part-II

Magnetic Field


Exercises, Q31 to Q40


    31.  A proton describes a circle of radius 1 cm in a magnetic field of strength 0·10 T. What would be the radius of the circle described by an α-particle moving with the same speed in the same magnetic field?      



ANSWER: The magnetic force on a moving charged particle is perpendicular to its motion, so it provides the centripetal force to move the particle in a circle. If m =mass, v =speed, r =radius of the circle, B =strength of the magnetic field and q =charge on the particle then,

 qvB = mv²/r   

→r =mv/qB

It shows that for constant v and B, the radius of the circle is directly proportional to m and inversely proportional to q. An alpha particle has mass four times the mass of a proton and the charge twice that of a proton. Hence the radius of the circle described by an alpha particle,

  R =4mv/2qB =2*mv/qB =2r

Given, r =1 cm, hence

R =2 cm.         





 

    32.  An electron having a kinetic energy of 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.      



ANSWER: Kinetic energy, K =½mv²,   

→v² =2K/m

Since, B =mv/qr  

→B² =m²*2K/mq²r² =2mK/q²r²

  =2mK/e²r²

Putting values,

B²=2*9.11X10⁻³¹*100*1.60x10⁻¹⁹/(1.60x10⁻¹⁹)²*(0.1)²

 =11.39X10⁻⁸

→B =3.4X10⁻⁴ T    

   Time period of the electron,   

T =2πr/v, so

Frequency = 1/T =v/2πr

 =√(2K/m)/2πr   

 =√(K/2π²mr²)

=√{100*1.6X10⁻¹⁹/2*π²*9.11x10⁻³¹*(0.1)²}

= √{89X10¹²}

= 9.4x10⁶            






    33.  Protons having kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.      



ANSWER: The proton will move in a circle and its radius is =l.   

We have seen in problem 32,  

B² =2mK/q²r²  

Here q =e, r =l, and let mass of a proton =mₒ then 

B² =2mₒK/e²l²     

→B =√(2mₒK)el            









    34.  A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1·0x10⁶ m/s. It is then injected perpendicularly into a magnetic field of strength 0·2 T. Find the radius of the circle described by it.       



ANSWER: Let the charge on the particle =q. In moving it through a potential difference of V, work done on it =qV. This work done is equal to the kinetic energy gained =½mv². So,  

qV =½mv²

→q =½mv²/V

We have the radius of the circle

r =mv/qB

 =mv*(2V/mv²)/B

 =2V/vB

Put values,

r =2*12000/(1x10⁶*0.2)

  = 12x10⁻² m

 = 12 cm.    







   

    35.  Doubly ionized helium ions are projected with a speed of 10 km/s in a direction perpendicular to a uniform magnetic field of magnitude 1·0 T. Find (a) the force acting on an ion. (b) the radius of the circle in which it circulates and (c) the time taken by an ion to complete the circle.      



ANSWER: (a) Given, q =2e, 

v =1.0X10⁴ m/s, B =1.0 T.

The force acting on an ion,

F =qvB

 =2*1.6X10⁻¹⁹*1.0X10⁴*1.0

 =3.2x10⁻¹⁵ N


(b) The radius of the circle in which it circulates,

r =mv/qB

The mass of a helium ion will be about four times the mass of a proton, so

r =4*1.67x10⁻²⁷*1.0x10⁴/(2*1.6x10⁻¹⁹)*1.0 m

  =2.1X10⁻⁴ m.  


(c) Time taken by an ion to complete the circle  

T =distance/speed

 =2πr/v

 =2*3.14*2.1x10⁻⁴/1.0x10⁴ s

 =1.31x10⁻⁷ s.                           








    36.  A proton is projected with a velocity of 3x10⁶ m/s perpendicular to a uniform magnetic field of 0·6 T. Find the acceleration of the proton.      



ANSWER: Given, v =3x10⁶ m/s, B =0.6 T. So the force on the proton perpendicular to its motion, 

 F = evB

   =1.6x10⁻¹⁹*3x10⁶*0.6 N

   =2.88x10⁻¹³ N

Acceleration of the proton in the direction of force,

a =F/m

 =2.88x10⁻¹³/1.67x10⁻²⁷ m/s²

 =1.72x10¹⁴ m/s²           









    37.  (a) An electron moves along a circle of radius 1 m in a perpendicular magnetic field of strength 0·50 T. What would be its speed? Is it reasonable?

    (b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its speed?       



ANSWER: (a) Given, r = 1 m,

 B =0.50 T 

We have radius of the circle

r =mv/qB  

→v =rqB/m

  =1*1.6x10⁻¹⁹*0.50/9.11x10⁻³¹

  =8.8x10¹⁰ m/s.

    It does not sound reasonable because the speed of light is 3x10⁸ m/s and according to the theory of relativity no matter should have a speed more than the speed of light.               

  

(b) The charge on a proton is the same as the electron but opposite in nature, but the mass is 1840 times more than the electron. So the mass of the proton,

m' =1840m  

Speed of proton in the same condition,

v' =rqB/1840m 
   =v/1840
   =8.8x10¹⁰/1840 m/s   
   =4.8x10⁷ m/s.        



                        





    38.  A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure (34-E13). (a) Find the radius of the circular arc it describes in the magnetic field. (b) Find the angle subtended by the arc at the center. (c) How long does the particle stay inside the magnetic field? (d) Solve the three parts of the above problem if the charge q on the particle is negative.      
Figure for Q-38


ANSWER: (a) The particle enters the magnetic field perpendicularly, hence the magnetic force at right angles to its movement on it,    

F =qvB,

This force acts as a centripetal force for the circular motion of the particle. If r is the radius of the circular arc, then,   

mv²/r =qvB

→r = mv/qB  


(b)  The direction of the velocity at the entrance will be tangent to the arc.
Diagram for Q-38b


 From geometry, the angle between the radius and the chord of the arc will also be equal to theta. In the isosceles triangle OAB, the angle subtended by the arc at center  =π -2θ  


(c) Distance traveled by the particle inside the magnetic field = length of the arc, S =(π -2θ)*2πr/2π 

   =(π -2θ)r

The time taken by the particle in traveling this distance = S/v

  =(π-2θ)r/(qBr/m)

 =(π -2θ)m/qB

{ Since v =qBr/m}


(d) (i) If the charge on the particle is negative, the radius of the circular arc will still be the same i.e.  

     

 r =mv/qB, only the movement will now be clockwise, opposite. 


(ii) As we see in the diagram, the angle between the chord and the radius is also θ. Hence the angle subtended by the smaller arc (that is outside the magnetic field and imaginary) at the center =π -2θ

Diagram for Q-38c


   So the angle subtended at the center by the larger arc (the actual path traveled by the particle) is, 

 =2π -(π- 2θ) 

 =π +2θ


(iii) The distance traveled by the particle =the length of the arc inside the magnetic field, which is equal to

 {(π +2θ)/2π}*2πr

={π+2θ}r

Time taken in traveling this distance

={π+2θ}r/v

=(π+2θ)m/qB

{Since r =mv/qB so r/v =m/qB}


 

                 



 

    39.  A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation (figure 34-E14) of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than

(a) mv/qB  (b) mv/2qB  (c) 2mv/qB. 
Figure for Q-39



ANSWER: (a) As we have seen, the radius of the arc inscribed by the particle inside the magnetic field is mv/qB. Since the width of the magnetic field, d is slightly less than mv/qB, the particle will inscribe only a quarter of a circle. In this case, the direction of the final speed will be perpendicular to the direction at entry.
Diagram for Q-39 (a)

   Hence the angle of deviation when the particle comes out of the magnetic field =π/2. 


(b) When d =mv/2qB =r/2.   

Let the angle subtended by the arc traversed by the particle = θ.

  From geometry as in the figure below,
Diagram for Q-39 (b)


Sinθ =(r/2)/r =1/2 =Sin π/6

θ =π/6

The angle of deviation here is the angle between the tangents at the entry and exit of the particle. It will be equal to the angle subtended by the chord at the center = θ =π/6.


(c) Since the width d of the magnetic field is more than the radius of the described circle, the particle will not exit from the right face, instead it will describe a semicircle and come out from the entry face. Since the velocities at entry and exit are just opposite, the angle deviation will be equal to π.

     



 

    40.  A narrow beam of singly charged carbon ions, moving at a constant velocity of 6·0x10⁴ m/s, is sent perpendicularly in a rectangular region having a uniform magnetic field B = 0·5 T (figure 34-E15). It is found that two beams emerge from the field in the backward direction, the separation from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1·6x10⁻²⁷) kg, where A is the mass number.      
The figure for Q-40


ANSWER: (a)  Given, B =0.50 T,

 v =6.0X10⁴ m/s,

 q =e =1.6X10⁻¹⁹ C,

for the first beam, r =3/2 cm =1.5 cm =0.015 m

Hence mass of each atom in the first emerging beam,

m = rqB/v 

   = reB/v

   =0.015*1.6X10⁻¹⁹*0.5/6x10⁴ kg

   =2.0x10⁻²⁶ kg

Dividing it by the mass of a proton,

m/mₒ =2.0x10⁻²⁶/1.67X10⁻²⁷ = 1.2*10

   =12

    So one atom of this carbon beam is twelve times heavier than a proton. A carbon atom has 6 protons, thus the other 6 particles are neutrons. It is a ¹²C isotope beam.   


(b) For the other emerging beam,

 r =3.5/2 cm =1.75 cm.   

m/mₒ =12*(1.75/1.5) = 14

   So this isotope has 6 protons and 8 neutrons in its nucleus, it is a ¹⁴C isotope beam.       

          

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